<?xml version="1.0" encoding="utf-8" standalone="yes"?><rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom"><channel><title>Dp on 111qqz's blog</title><link>https://111qqz.com/en/tags/dp/</link><description>Recent content in Dp on 111qqz's blog</description><generator>Hugo -- gohugo.io</generator><language>en</language><copyright>© 2015-2026 111qqz</copyright><lastBuildDate>Sat, 11 Nov 2017 07:11:25 +0000</lastBuildDate><atom:link href="https://111qqz.com/en/tags/dp/index.xml" rel="self" type="application/rss+xml"/><item><title>SPOJ LCS2 Longest Common Substring 2[后缀自动机+dp]</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-11-11-spoj-lcs2/</link><pubDate>Sat, 11 Nov 2017 07:11:25 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-11-11-spoj-lcs2/</guid><description>&lt;h1 class="relative group"&gt;题意：
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&lt;p&gt;求n个串的最长公共子串，n&amp;lt;=10&lt;/p&gt;</description></item><item><title>poj 3249 Test for Job (拓扑排序+dp)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-11-07-poj-3249/</link><pubDate>Tue, 07 Nov 2017 06:13:22 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-11-07-poj-3249/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=3249" target="_blank" rel="noreferrer"&gt;http://poj.org/problem?id=3249&lt;/a&gt;&lt;/p&gt;

&lt;h1 class="relative group"&gt;题意：
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&lt;p&gt;给一个DAG,现要从一条入度为0的点到一个出度为0的点，问最大点权和。&lt;/p&gt;</description></item><item><title>codeforces 855 B. Marvolo Gaunt's Ring (前缀最大，dp)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-09-25-codeforces-855-b/</link><pubDate>Mon, 25 Sep 2017 13:12:25 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-09-25-codeforces-855-b/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/contest/855/problem/B" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出n,p,q,r,以及n（1E5）个数，所有数的范围都是[-1E9,1E9],现在问p_a[i]+q_a[j]+r*a[k]的最大值，满足1&amp;lt;=i&amp;lt;=j&amp;lt;=k&amp;lt;=n&lt;/p&gt;</description></item><item><title>leetcode 152. Maximum Product Subarray (最大连续子序列乘积，dp)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-04-14-leetcode-152-maximum-product-subarray/</link><pubDate>Fri, 14 Apr 2017 11:33:30 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-04-14-leetcode-152-maximum-product-subarray/</guid><description>&lt;p&gt;Find the contiguous subarray within an array (containing at least one number) which has the largest product.&lt;/p&gt;
&lt;p&gt;For example, given the array &lt;code&gt;[2,3,-2,4]&lt;/code&gt;,
the contiguous subarray &lt;code&gt;[2,3]&lt;/code&gt; has the largest product = &lt;code&gt;6&lt;/code&gt;.&lt;/p&gt;</description></item><item><title>leetocde 63. Unique Paths II</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-04-11-63-unique-paths-ii/</link><pubDate>Tue, 11 Apr 2017 10:50:57 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-04-11-63-unique-paths-ii/</guid><description>&lt;p&gt;Follow up for &amp;ldquo;Unique Paths&amp;rdquo;:&lt;/p&gt;
&lt;p&gt;Now consider if some obstacles are added to the grids. How many unique paths would there be?&lt;/p&gt;
&lt;p&gt;An obstacle and empty space is marked as &lt;code&gt;1&lt;/code&gt; and &lt;code&gt;0&lt;/code&gt; respectively in the grid.&lt;/p&gt;</description></item><item><title>leetcode 64. Minimum Path Sum (二维dp)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-04-10-leetcode-64-minimum-path-sum/</link><pubDate>Mon, 10 Apr 2017 02:35:20 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-04-10-leetcode-64-minimum-path-sum/</guid><description>&lt;p&gt;Given a &lt;em&gt;m&lt;/em&gt; x &lt;em&gt;n&lt;/em&gt; grid filled with non-negative numbers, find a path from top left to bottom right which &lt;em&gt;minimizes&lt;/em&gt; the sum of all numbers along its path.&lt;/p&gt;
&lt;p&gt;&lt;strong&gt;Note:&lt;/strong&gt; You can only move either down or right at any point in time.&lt;/p&gt;</description></item><item><title>BZOJ 2748: [HAOI2012]音量调节 (dp)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-04-02-bzoj-2748/</link><pubDate>Sun, 02 Apr 2017 06:50:50 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-04-02-bzoj-2748/</guid><description>&lt;h2 class="relative group"&gt;2748: [HAOI2012]音量调节
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&lt;p&gt;Time Limit: 3 Sec  Memory Limit: 128 MB
Submit: 1814  Solved: 1148
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=2748" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=2748" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=2748" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>BZOJ 1207: [HNOI2004]打鼹鼠 (LIS)</title><link>https://111qqz.com/en/post/acm-icpc/2017/2017-01-26-bzoj-1207-hnoi2004e68993e9bcb9e9bca0-lis/</link><pubDate>Thu, 26 Jan 2017 08:12:38 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2017/2017-01-26-bzoj-1207-hnoi2004e68993e9bcb9e9bca0-lis/</guid><description>&lt;h2 class="relative group"&gt;1207: [HNOI2004]打鼹鼠
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&lt;p&gt;Time Limit: 10 Sec  Memory Limit: 162 MB
Submit: 2854  Solved: 1390
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1207" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1207" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1207" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>(dp专题006)hdu 2602 Bone Collector（01背包）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-16-hdu2602/</link><pubDate>Wed, 16 Nov 2016 07:28:33 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-16-hdu2602/</guid><description>&lt;p&gt;题目链接&lt;/p&gt;
&lt;p&gt;题意:容量为V的背包，n个骨头，给出价值和体积，问最多能装多少价值的背包。&lt;/p&gt;</description></item><item><title>[dp专题005]hdu 1864最大报销额（01背包，垃圾题）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-16-hdu1864/</link><pubDate>Wed, 16 Nov 2016 07:04:20 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-16-hdu1864/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=1864" target="_blank" rel="noreferrer"&gt;hdu1864题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：中文题目，不多说了。&lt;/p&gt;
&lt;p&gt;思路：正解是01背包，呵呵呵。&lt;/p&gt;
&lt;p&gt;出题人是傻逼吗？&lt;/p&gt;</description></item><item><title>(dp专题004)hdu 2955Robberies（01背包变形）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-15-hdu2955/</link><pubDate>Tue, 15 Nov 2016 11:19:56 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-15-hdu2955/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2955" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意： 给出n个银行 ，以及抢劫每个银行可以得到的价值和被抓的概率，不同银行之间被抓的概率是相互独立的，现在给出安全概率p，只有当概率从小于安全概率时才是安全的，问最多能抢劫多少价值。&lt;/p&gt;</description></item><item><title>(dp专题003)hdu 4055 Number String(dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-13-hdu-4055/</link><pubDate>Sun, 13 Nov 2016 14:38:06 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-13-hdu-4055/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4055" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出n(n&amp;lt;=1E3)个字符，字符可能为&amp;rsquo;D&amp;rsquo;,&amp;lsquo;I&amp;rsquo;,&amp;rsquo;?&amp;rsquo;，第i位对应的字符分别表示，第i位大于第i+1位，第i位小于第i+1位，或者不确定。&lt;/p&gt;</description></item><item><title>【dp专题002】hdu 4489 The King’s Ups and Downs (dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-13-e38090dpe4b893e9a298002e38091hdu-4489-the-kings-ups-and-downs-dp/</link><pubDate>Sun, 13 Nov 2016 11:33:20 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-13-e38090dpe4b893e9a298002e38091hdu-4489-the-kings-ups-and-downs-dp/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4489" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：问长度为n的“波浪”型排列（即1..n每个数出现一次）有多少。波浪型的含义是，“高低高”或者“低高低”&lt;/p&gt;</description></item><item><title>【dp专题001】bzoj 1009: [HNOI2008]GT考试 (字符串上dp+kmp+矩阵加速线性递推式)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-13-bzoj-1009/</link><pubDate>Sun, 13 Nov 2016 07:21:02 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-13-bzoj-1009/</guid><description>&lt;h2 class="relative group"&gt;1009: [HNOI2008]GT考试
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&lt;p&gt;Time Limit: 1 Sec  Memory Limit: 162 MB
Submit: 3127  Solved: 1926
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1009" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1009" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1009" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>[dp专题000]uva 10328 Coin Toss (java 大数+dp)（Unsolved）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-11-12-uva10328/</link><pubDate>Sat, 12 Nov 2016 12:39:25 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-11-12-uva10328/</guid><description>&lt;p&gt;题目链接&lt;/p&gt;
&lt;p&gt;题意：问长度为n，每个位置由且仅有‘H’和&amp;rsquo;T&amp;rsquo;组成的序列中，至少有连续k个‘H’出现的方案数。&lt;/p&gt;</description></item><item><title>codeforces 605 A. Sorting Railway Cars (dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-10-04-cf605a/</link><pubDate>Tue, 04 Oct 2016 12:59:16 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-10-04-cf605a/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/problemset/problem/605/A" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出一个n个数的排列，每次可以把一个数放到最前面或者最后面的位置，问至少要进行多少次操作才能使得数列升序。&lt;/p&gt;</description></item><item><title>hdu 5904 LCIS (dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-09-25-hdu-5904-lcis-dp/</link><pubDate>Sun, 25 Sep 2016 20:13:38 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-09-25-hdu-5904-lcis-dp/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=5904" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：
给定两个序列，求它们的最长公共递增子序列的长度, 并且这个子序列的值是连续的
思路：以值为连续做入手点。&lt;/p&gt;</description></item><item><title>斜率优化学习笔记</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-09-24-e6969ce78e87e4bc98e58c96e5ada6e4b9a0e7ac94e8aeb0/</link><pubDate>Sat, 24 Sep 2016 15:38:00 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-09-24-e6969ce78e87e4bc98e58c96e5ada6e4b9a0e7ac94e8aeb0/</guid><description>&lt;p&gt;&lt;a href="http://www.cnblogs.com/ka200812/archive/2012/08/03/2621345.html" target="_blank" rel="noreferrer"&gt;参考博客&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;这个东西英文好像叫做：convex hull trick&lt;/p&gt;
&lt;p&gt;&lt;a href="http://wcipeg.com/wiki/Convex_hull_trick" target="_blank" rel="noreferrer"&gt;Convex_hull_trick_wiki&lt;/a&gt;
&lt;a href="http://codeforces.com/blog/entry/11339" target="_blank" rel="noreferrer"&gt;codeforces convex hull trick&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;简单说说我的理解：斜率优化是一种数形结合的思想。。。&lt;/p&gt;</description></item><item><title>hdu 5763 || 2016 multi #4 1001 Another Meaning (kmp+dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-08-11-hdu-5763/</link><pubDate>Thu, 11 Aug 2016 15:41:57 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-08-11-hdu-5763/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=5763" target="_blank" rel="noreferrer"&gt;hdu 5763 题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给定两个字符串A和B，每个出现在A中的B(可以overlap)都有两种含义，问A串一共可能有多少种含义。&lt;/p&gt;</description></item><item><title>codeforces 429 B. Working out (dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-27-cf429b/</link><pubDate>Wed, 27 Jul 2016 19:05:42 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-27-cf429b/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/problemset/problem/429/B" target="_blank" rel="noreferrer"&gt;cf429 b 题目链接&lt;/a&gt;
题意：&lt;/p&gt;
&lt;p&gt;n*m个格子，每个格子有一个人value a[i][j]&amp;gt;0，连个人，一个从左上角到右下角，每次只能向下或者向右移动，一个从左下到右上，每次只能向上或者向右移动，现在要求两个人恰好相遇一次，相遇点的a不算数，问在满足这样的条件下使得两个人的a最大。。。&lt;strong&gt;（很坑的一点是。。这里相遇并不考虑时间。。就是说，不在同一时间都到达过某一格子，也认为相遇。所以我认为，题目含义更准确的说法是，路径只有一个交点）&lt;/strong&gt;&lt;/p&gt;</description></item><item><title>hdu 2018 母牛的故事 (基础dp，记忆化搜索)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-27-hdu-2018/</link><pubDate>Wed, 27 Jul 2016 05:57:39 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-27-hdu-2018/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2018" target="_blank" rel="noreferrer"&gt;hdu2018题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意:第1年有1头奶牛，每年生一头奶牛，新生的奶牛从生下来的第四年（包括生下来那年）也开始每年一头奶牛。
问第n年有多少头奶牛。
思路：最容易想到的。。递推一下。。。dp[i] = dp[i-1] + dp[i-3] (&lt;strong&gt;注意初始化不是一个dp[1]=1,而是dp[1..4]=1..4&lt;/strong&gt;)&lt;/p&gt;</description></item><item><title>hdu 2084 数塔 (基础dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-27-hdu-2084/</link><pubDate>Wed, 27 Jul 2016 05:12:10 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-27-hdu-2084/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2084" target="_blank" rel="noreferrer"&gt;hdu2084题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：dp入门题。。。数字三角形。。&lt;/p&gt;
&lt;p&gt;思路：&lt;/p&gt;
&lt;p&gt;昨天看mit公开课。。。讲到dp的精髓是sub-problem+ reuse&amp;hellip;&lt;/p&gt;</description></item><item><title>hdu 4283 You Are the One (区间dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-26-hdu-4283/</link><pubDate>Tue, 26 Jul 2016 07:38:44 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-26-hdu-4283/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4283" target="_blank" rel="noreferrer"&gt;hdu 4283题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：有N个人按顺序排成一排上台表演，每个都有一个num[]值，若在他是第k个上场的人，则会有num[]*(k-1)的unhappiness。台下有一个黑屋（stack），对每一个人，可以选择让他先进屋子或者直接上台。现在让你找到一个最优方案使得所有人的unhappiness之和最小。&lt;/p&gt;</description></item><item><title>poj 3661 Running (区间dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj-3661-running-e58cbae997b4dp/</link><pubDate>Mon, 25 Jul 2016 16:51:22 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj-3661-running-e58cbae997b4dp/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=3661" target="_blank" rel="noreferrer"&gt;poj 3661题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：锻炼，一共n分钟，每分钟可以选择跑步或者休息，第i分钟跑步可以跑d[i]米，并增加一点疲劳度，如果选择休息，那么每分钟减少1点疲劳值。一旦开始休息，必须休息到疲劳值为0才能再次开始跑步。疲劳值不能超过m.第n分钟的时候疲劳值必须为0，否则之后会感觉身体被掏空。问n分钟最远多多远。&lt;/p&gt;</description></item><item><title>poj 1651 Multiplication Puzzle (区间dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj-1651/</link><pubDate>Mon, 25 Jul 2016 15:09:40 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj-1651/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=1651" target="_blank" rel="noreferrer"&gt;poj 1651题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：n个数，删掉a[i]的得分是a[i]*a[i-1]*a[i+1]，两个端点的不允许删。问删完n-2个数得到的最小分数是多少。&lt;/p&gt;</description></item><item><title>poj 3280 Cheapest Palindrome (区间dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj-3280/</link><pubDate>Mon, 25 Jul 2016 13:26:14 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj-3280/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=3280" target="_blank" rel="noreferrer"&gt;poj 3280 题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：一个字符串，给出添加一个字符或者删掉该字符的花费，问最小的话费使得字符串变成回文串。&lt;/p&gt;</description></item><item><title>light oj 1422 - Halloween Costumes (区间dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-light-oj-1422-halloween-costumes-e58cbae997b4dp/</link><pubDate>Mon, 25 Jul 2016 11:24:27 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-light-oj-1422-halloween-costumes-e58cbae997b4dp/</guid><description>&lt;p&gt;&lt;a href="http://lightoj.com/volume_showproblem.php?problem=1422" target="_blank" rel="noreferrer"&gt;light oj 1422 题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：&lt;/p&gt;
&lt;p&gt;按顺序去参加舞会。每个舞会对衣服都有要求。可以连续穿好多件衣服。需要时候就脱下来，但是一旦脱下来，这件衣服就报废了。问最少需要几件衣服。&lt;/p&gt;</description></item><item><title>poj 2955 Brackets（区间dp....括号匹配。。。人生第一道区间dp）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj2955/</link><pubDate>Mon, 25 Jul 2016 07:46:23 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-07-25-poj2955/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=2955" target="_blank" rel="noreferrer"&gt;poj2955题目链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;题意：给出若干括号，问最大匹配数是多少。&lt;/p&gt;
&lt;p&gt;思路：没有思路。我知道这是dp。。。然后其他就什么都不知道了。。。转移方程？ 完全没思路。。知道了转移方程。。。。嗯，还是不会。。。边界怎么写？状态怎么推？循环顺序? 循环次序？我一点思路都没有。。。。。&lt;/p&gt;</description></item><item><title>BZOJ 1649: [Usaco2006 Dec]Cow Roller Coaster (dp，类似01背包)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-11-bzoj-1649/</link><pubDate>Mon, 11 Apr 2016 09:19:52 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-11-bzoj-1649/</guid><description>&lt;h2 class="relative group"&gt;
 &lt;div id="" class="anchor"&gt;&lt;/div&gt;
 
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&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 504  Solved: 265
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1649" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1649" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1649" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;

&lt;h2 class="relative group"&gt;Description
 &lt;div id="description" class="anchor"&gt;&lt;/div&gt;
 
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&lt;/h2&gt;
&lt;p&gt;The cows are building a roller coaster! They want your help to design as fun a roller coaster as possible, while keeping to the budget. The roller coaster will be built on a long linear stretch of land of length L (1 &amp;lt;= L &amp;lt;= 1,000). The roller coaster comprises a collection of some of the N (1 &amp;lt;= N &amp;lt;= 10,000) different interchangable components. Each component i has a fixed length Wi (1 &amp;lt;= Wi &amp;lt;= L). Due to varying terrain, each component i can be only built starting at location Xi (0 &amp;lt;= Xi &amp;lt;= L-Wi). The cows want to string together various roller coaster components starting at 0 and ending at L so that the end of each component (except the last) is the start of the next component. Each component i has a &amp;ldquo;fun rating&amp;rdquo; Fi (1 &amp;lt;= Fi &amp;lt;= 1,000,000) and a cost Ci (1 &amp;lt;= Ci &amp;lt;= 1000). The total fun of the roller coster is the sum of the fun from each component used; the total cost is likewise the sum of the costs of each component used. The cows&amp;rsquo; total budget is B (1 &amp;lt;= B &amp;lt;= 1000). Help the cows determine the most fun roller coaster that they can build with their budget.&lt;/p&gt;</description></item><item><title>BZOJ 1644: [Usaco2007 Oct]Obstacle Course 障碍训练课 (BFS,DP)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-10-bzoj-1644-usaco2007-octobstacle-course-e99a9ce7a28de8aeade7bb83e8afbe-bfsdp/</link><pubDate>Sun, 10 Apr 2016 12:41:46 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-10-bzoj-1644-usaco2007-octobstacle-course-e99a9ce7a28de8aeade7bb83e8afbe-bfsdp/</guid><description>&lt;h2 class="relative group"&gt;1644: [Usaco2007 Oct]Obstacle Course 障碍训练课
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&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 451  Solved: 226
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1644" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1644" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1644" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>BZOJ 1632: [Usaco2007 Feb]Lilypad Pond (BFS,dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-06-bzoj-1632/</link><pubDate>Wed, 06 Apr 2016 15:37:13 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-06-bzoj-1632/</guid><description>&lt;h2 class="relative group"&gt;1632: [Usaco2007 Feb]Lilypad Pond
 &lt;div id="1632-usaco2007-feblilypad-pond" class="anchor"&gt;&lt;/div&gt;
 
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&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 496  Solved: 153
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1632" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1632" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1632" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>BZOJ 1618: [Usaco2008 Nov]Buying Hay 购买干草 (完全背包)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-03-bzoj-1618-usaco2008-novbuying-hay-e8b4ade4b9b0e5b9b2e88d89-e5ae8ce585a8e8838ce58c85/</link><pubDate>Sun, 03 Apr 2016 11:53:05 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-03-bzoj-1618-usaco2008-novbuying-hay-e8b4ade4b9b0e5b9b2e88d89-e5ae8ce585a8e8838ce58c85/</guid><description>&lt;h2 class="relative group"&gt;1618: [Usaco2008 Nov]Buying Hay 购买干草
 &lt;div id="1618-usaco2008-novbuying-hay-购买干草" class="anchor"&gt;&lt;/div&gt;
 
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 &lt;/span&gt;
 
&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 906  Solved: 456
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1618" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1618" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1618" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>BZOJ 1617: [Usaco2008 Mar]River Crossing渡河问题 (DP)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-03-bzoj1617/</link><pubDate>Sun, 03 Apr 2016 11:34:42 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-03-bzoj1617/</guid><description>&lt;h2 class="relative group"&gt;1617: [Usaco2008 Mar]River Crossing渡河问题
 &lt;div id="1617-usaco2008-marriver-crossing渡河问题" class="anchor"&gt;&lt;/div&gt;
 
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 &lt;/span&gt;
 
&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 837  Solved: 606
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1617" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1617" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1617" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>BZOJ 1616: [Usaco2008 Mar]Cow Travelling游荡的奶牛(DP)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-03-bzoj1616/</link><pubDate>Sun, 03 Apr 2016 07:52:18 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-03-bzoj1616/</guid><description>&lt;h2 class="relative group"&gt;1616: [Usaco2008 Mar]Cow Travelling游荡的奶牛
 &lt;div id="1616-usaco2008-marcow-travelling游荡的奶牛" class="anchor"&gt;&lt;/div&gt;
 
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 class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"&gt;
 &lt;a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#1616-usaco2008-marcow-travelling%e6%b8%b8%e8%8d%a1%e7%9a%84%e5%a5%b6%e7%89%9b" aria-label="Anchor"&gt;#&lt;/a&gt;
 &lt;/span&gt;
 
&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 1012  Solved: 553
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1616" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1616" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1616" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>BZOJ 1613: [Usaco2007 Jan]Running贝茜的晨练计划 (dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-02-bzoj1613/</link><pubDate>Sat, 02 Apr 2016 09:41:36 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-02-bzoj1613/</guid><description>&lt;h2 class="relative group"&gt;1613: [Usaco2007 Jan]Running贝茜的晨练计划
 &lt;div id="1613-usaco2007-janrunning贝茜的晨练计划" class="anchor"&gt;&lt;/div&gt;
 
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 class="absolute top-0 w-6 transition-opacity opacity-0 -start-6 not-prose group-hover:opacity-100 select-none"&gt;
 &lt;a class="text-primary-300 dark:text-neutral-700 !no-underline" href="#1613-usaco2007-janrunning%e8%b4%9d%e8%8c%9c%e7%9a%84%e6%99%a8%e7%bb%83%e8%ae%a1%e5%88%92" aria-label="Anchor"&gt;#&lt;/a&gt;
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&lt;/h2&gt;
&lt;p&gt;Time Limit: 5 Sec  Memory Limit: 64 MB
Submit: 1468  Solved: 706
[&lt;a href="http://www.lydsy.com/JudgeOnline/submitpage.php?id=1613" target="_blank" rel="noreferrer"&gt;Submit&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/problemstatus.php?id=1613" target="_blank" rel="noreferrer"&gt;Status&lt;/a&gt;][&lt;a href="http://www.lydsy.com/JudgeOnline/bbs.php?id=1613" target="_blank" rel="noreferrer"&gt;Discuss&lt;/a&gt;]&lt;/p&gt;</description></item><item><title>最长上升子序列nlogn解法</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-04-01-e69c80e995bfe4b88ae58d87e5ad90e5ba8fe58897nlogne8a7a3e6b395/</link><pubDate>Fri, 01 Apr 2016 12:15:41 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-04-01-e69c80e995bfe4b88ae58d87e5ad90e5ba8fe58897nlogne8a7a3e6b395/</guid><description>&lt;p&gt;首先回顾一下n^2的做法。
状态转移方程为dp[i] =max(1,dp[j]) (1=&amp;lt;j&amp;lt;=i-1&amp;amp;&amp;amp;a[i]&amp;gt;a[j])&lt;/p&gt;
&lt;div class="highlight-wrapper"&gt;&lt;div class="highlight"&gt;&lt;pre tabindex="0" class="chroma"&gt;&lt;code class="language-cpp" data-lang="cpp"&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 1&lt;/span&gt;&lt;span class="cl"&gt;&lt;span class="k"&gt;for&lt;/span&gt; &lt;span class="p"&gt;(&lt;/span&gt; &lt;span class="kt"&gt;int&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="mi"&gt;1&lt;/span&gt; &lt;span class="p"&gt;;&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt; &lt;span class="o"&gt;&amp;lt;=&lt;/span&gt; &lt;span class="n"&gt;n&lt;/span&gt; &lt;span class="p"&gt;;&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="o"&gt;++&lt;/span&gt;&lt;span class="p"&gt;)&lt;/span&gt; &lt;span class="n"&gt;cin&lt;/span&gt;&lt;span class="o"&gt;&amp;gt;&amp;gt;&lt;/span&gt;&lt;span class="n"&gt;a&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="p"&gt;];&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 2&lt;/span&gt;&lt;span class="cl"&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 3&lt;/span&gt;&lt;span class="cl"&gt;	&lt;span class="k"&gt;for&lt;/span&gt; &lt;span class="p"&gt;(&lt;/span&gt; &lt;span class="kt"&gt;int&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="mi"&gt;1&lt;/span&gt; &lt;span class="p"&gt;;&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt; &lt;span class="o"&gt;&amp;lt;=&lt;/span&gt; &lt;span class="n"&gt;n&lt;/span&gt; &lt;span class="p"&gt;;&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="o"&gt;++&lt;/span&gt;&lt;span class="p"&gt;)&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 4&lt;/span&gt;&lt;span class="cl"&gt;	&lt;span class="p"&gt;{&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 5&lt;/span&gt;&lt;span class="cl"&gt;	 &lt;span class="n"&gt;dp&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="p"&gt;]&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="mi"&gt;1&lt;/span&gt;&lt;span class="p"&gt;;&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 6&lt;/span&gt;&lt;span class="cl"&gt;	 &lt;span class="k"&gt;for&lt;/span&gt; &lt;span class="p"&gt;(&lt;/span&gt; &lt;span class="kt"&gt;int&lt;/span&gt; &lt;span class="n"&gt;j&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="mi"&gt;1&lt;/span&gt; &lt;span class="p"&gt;;&lt;/span&gt; &lt;span class="n"&gt;j&lt;/span&gt; &lt;span class="o"&gt;&amp;lt;&lt;/span&gt; &lt;span class="n"&gt;i&lt;/span&gt; &lt;span class="p"&gt;;&lt;/span&gt; &lt;span class="n"&gt;j&lt;/span&gt;&lt;span class="o"&gt;++&lt;/span&gt;&lt;span class="p"&gt;)&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 7&lt;/span&gt;&lt;span class="cl"&gt;	 &lt;span class="p"&gt;{&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 8&lt;/span&gt;&lt;span class="cl"&gt;		&lt;span class="k"&gt;if&lt;/span&gt; &lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="n"&gt;a&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="p"&gt;]&lt;/span&gt;&lt;span class="o"&gt;&amp;gt;&lt;/span&gt;&lt;span class="n"&gt;a&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;j&lt;/span&gt;&lt;span class="p"&gt;])&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt; 9&lt;/span&gt;&lt;span class="cl"&gt;		 &lt;span class="n"&gt;dp&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="p"&gt;]&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="n"&gt;max&lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="n"&gt;dp&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="p"&gt;],&lt;/span&gt;&lt;span class="n"&gt;dp&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;j&lt;/span&gt;&lt;span class="p"&gt;]&lt;/span&gt;&lt;span class="o"&gt;+&lt;/span&gt;&lt;span class="mi"&gt;1&lt;/span&gt;&lt;span class="p"&gt;);&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt;10&lt;/span&gt;&lt;span class="cl"&gt;	 &lt;span class="p"&gt;}&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt;11&lt;/span&gt;&lt;span class="cl"&gt; &lt;span class="n"&gt;ans&lt;/span&gt; &lt;span class="o"&gt;=&lt;/span&gt; &lt;span class="n"&gt;max&lt;/span&gt;&lt;span class="p"&gt;(&lt;/span&gt;&lt;span class="n"&gt;ans&lt;/span&gt;&lt;span class="p"&gt;,&lt;/span&gt;&lt;span class="n"&gt;dp&lt;/span&gt;&lt;span class="p"&gt;[&lt;/span&gt;&lt;span class="n"&gt;i&lt;/span&gt;&lt;span class="p"&gt;]);&lt;/span&gt;
&lt;/span&gt;&lt;/span&gt;&lt;span class="line"&gt;&lt;span class="ln"&gt;12&lt;/span&gt;&lt;span class="cl"&gt;	&lt;span class="p"&gt;}&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/code&gt;&lt;/pre&gt;&lt;/div&gt;&lt;/div&gt;
&lt;p&gt;然后我们发现，使得dp[i]得到同一个值的dp[j]可能有多个，那么选择哪个呢？
假设 x&amp;lt;y&amp;lt;i，a[x]&amp;lt;a[y],dp[x]==dp[y]，那么我们选择x好还是y好呢？
显然是x好。为什么？因为选择x潜力大。因为可能在x,y之间存在一个z,满足a[x]&amp;lt;a[z]&amp;lt;a[y],如果选择a[y]，就没有办法选择可能使长度更长的a[z]了。通俗得说。。我们要求的是最长上升子序列。。你一开始就弄那么大。。。后面还上哪上升去啊。。。长度小啊。。。&lt;/p&gt;</description></item><item><title>codeforces #120 div 2 (Virtual Participation)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-24-codeforces-120-div-2-virtual-participation/</link><pubDate>Thu, 24 Mar 2016 06:41:11 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-24-codeforces-120-div-2-virtual-participation/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/contest/190" target="_blank" rel="noreferrer"&gt;比赛链接&lt;/a&gt;&lt;/p&gt;
&lt;p&gt;两题QAQ&lt;/p&gt;
&lt;p&gt;A：7分钟1A 有n个大人m个小孩乘公交车，票价每人一元，一个大人最多免费带一个小孩，没有大人陪同的小孩不能乘车。 问是否有解，如果有解输出所有乘客付的钱的可能的最小值和可能的最大值。&lt;/p&gt;</description></item><item><title>codeforces #338 div2 B || 615B Longtail Hedgehog</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-22-cf615b/</link><pubDate>Tue, 22 Mar 2016 10:28:04 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-22-cf615b/</guid><description>&lt;p&gt;&lt;a href="http://codeforces.com/contest/615/problem/B" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：给出n个点，m条边，定义一条路径的价值为【路径长度*(路径终点的度)】，求最大价值。
思路：一月份的时候写过一个回溯。。。TLE22了。。。其实也能猜到是dp..但是无奈不会写。然而其实真的不难==
我们枚举路径的终点，dp[i]表示以点i为终点能得到的最长路径长度。&lt;/p&gt;</description></item><item><title>ural 1057. Amount of Degrees (b进制数位dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-18-ural1057/</link><pubDate>Fri, 18 Mar 2016 07:06:05 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-18-ural1057/</guid><description>&lt;p&gt;&lt;a href="http://acm.timus.ru/problem.aspx?space=1&amp;amp;num=1057" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：设条件A为一个数&lt;strong&gt;恰好&lt;/strong&gt;是k个&lt;strong&gt;互不相同&lt;/strong&gt;的b的整数次幂的和，问某一个区间内满足条件A的数的个数是有多少个。&lt;/p&gt;</description></item><item><title>hdu 4734 F(x) (数位dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-18-hdu4734/</link><pubDate>Fri, 18 Mar 2016 01:32:06 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-18-hdu4734/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4734" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;s
题意：将一个10进制数x按照2进制展开后得到的值设为f(x)&amp;hellip;现在给出a,b（10^9）问【0，b】中满足f[x]&amp;lt;=f[a]的数的个数。
思路：先算出f[a]&amp;hellip;然后我们发现f(x)最大也就10*2^10=10240.。。数组可以存下。。搞之。&lt;strong&gt;和上一道题类似。。我们不关心两个f函数的值具体是多少。。。只关心他们的相对大小情况。。所以还是可以合并成一个变量。。。然而我用两个变量为什么错了。。不懂==&lt;/strong&gt;&lt;/p&gt;</description></item><item><title>bc #75 C || hdu 5642 King's Order （数位dp）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-hdu5642/</link><pubDate>Thu, 17 Mar 2016 12:35:47 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-hdu5642/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=5642" target="_blank" rel="noreferrer"&gt;hdu5642题目链接&lt;/a&gt;
题意：问长度为n的仅由26个小写字母组成的合法字符串有多少个。如果某个字符连续出现四次或以上，则这个字符串为非法。否则为合法。&lt;/p&gt;</description></item><item><title>hdu 3709 Balanced Number (数位dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-hdu-3709-balanced-number-e695b0e4bd8ddp/</link><pubDate>Thu, 17 Mar 2016 11:46:31 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-hdu-3709-balanced-number-e695b0e4bd8ddp/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=3709" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：找到某区间中平衡数的个数。所谓平衡数是指，存在某个位置，值得两边的力矩相等。举个例子。。比如14326，如果把4作为中间。。那么左边=1&lt;em&gt;1=1 右边=3&lt;/em&gt;1+2&lt;em&gt;2+6&lt;/em&gt;2=19。。。
思路：枚举中间的pivot。。。注意如果是个位数也是平衡数（就是认为两边的力矩都是0了。。。），所以每一个位置都可能是平衡位置。。枚举的时候从1到len&amp;hellip;
一开始我是分别记录两边的值。。非常浪费空间。。。然而发现其实没必要。。&lt;strong&gt;我们只关心左边是否相等。。而不关心左右的值到底是多少。。所以可以把两边的值带符号合并成一个值&lt;/strong&gt;（pivot左边为+，pivot右边为负）。。。如果最后为0。。说明左右相等。。。&lt;/p&gt;</description></item><item><title>poj3252 Round Numbers (不允许前导0的二进制数位dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-poj3252/</link><pubDate>Thu, 17 Mar 2016 08:51:14 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-poj3252/</guid><description>&lt;p&gt;&lt;a href="http://poj.org/problem?id=3252" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：问某区间中，round number 的个数是多少。所谓round number,当且仅当一个数的二进制表示中，‘0’的个数大于等于‘1’的个数。
思路：简单数位dp..和windy数那道题类似，都是不允许前导0.。。所以在dfs中要加一维判断前面是否有非0的数。。。&lt;/p&gt;</description></item><item><title>hdu 4507 吉哥系列故事——恨7不成妻 (返回平方和的数位dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-hdu4507/</link><pubDate>Thu, 17 Mar 2016 07:55:35 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-17-hdu4507/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4507" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：如果一个整数符合下面3个条件之一，那么我们就说这个整数和7有关——
1、整数中某一位是7；
2、整数的每一位加起来的和是7的整数倍；
3、这个整数是7的整数倍；&lt;/p&gt;</description></item><item><title>hdu 3652 B-number (带整除的数位dp )</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-16-hdu3652/</link><pubDate>Wed, 16 Mar 2016 02:23:08 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-16-hdu3652/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=3652" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：给出n,问[1,n]中，满足包含“13”且这个数（不是各位的和）能被13整除的数的个数。
思路：依然是数位dp..不过有一个小tip。。&lt;/p&gt;</description></item><item><title>hdu 4722 good numbers (带整除的数位dp)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-16-hdu4722/</link><pubDate>Wed, 16 Mar 2016 01:17:27 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-16-hdu4722/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=4722" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：求一个区间内所有位数字之和能被10整除的数的个数。
思路：数位dp，dfs要一个参数记录从最高位到现在的pos位置的数字之和的结果。&lt;/p&gt;</description></item><item><title>bzoj 1026 windy数(数位dp入门题)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-15-bzoj1026/</link><pubDate>Tue, 15 Mar 2016 13:05:13 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-15-bzoj1026/</guid><description>&lt;p&gt;&lt;a href="http://www.lydsy.com/JudgeOnline/problem.php?id=1026" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意：不含前导零且相邻两个数字之差至少为2的正整数被称为windy数。 windy想知道，在A和B之间，包括A和B，总共有多少个windy数？
思路：数位dp
这道题的特点是前面不允许前导0，也就是说，如果第i位前面全是0的话，这个数就变成了i位数，i就变成了最高位，而最高位没有前面的数（**如果这里不考虑不允许前导0这个因素而把前面的一个数认为成是0就错了） **最高位的数可以直接取。 还有记忆化调用以及存储的时候也要注意&amp;hellip;只有当位数相同的时候转移才有意义。 具体的方法是dfs中多了一个prehasnonzero的bool变量，就是字面意思，判断当前位置前面的位置是够存在一个非0的值。&lt;/p&gt;</description></item><item><title>hdu 3555 Bomb （数位dp入门题）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-15-hdu3555/</link><pubDate>Tue, 15 Mar 2016 11:48:19 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-15-hdu3555/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=3555" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
题意:问从1到n的所有数中，有多少个数含有数字串“49”
思路：和上一道不要62很像，但是由于是要统计有49的，但是有49的情况实在太多了，正难则反，用减法定理反过来考虑，先统计出不含49的数的个数，这样就和不要62一样了，然后再用总数减。&lt;/p&gt;</description></item><item><title>hdu 2089 不要62 （数位dp模板题，附带详细解释）</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-03-15-hdu-2089/</link><pubDate>Tue, 15 Mar 2016 11:27:43 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-03-15-hdu-2089/</guid><description>&lt;p&gt;&lt;a href="http://acm.hdu.edu.cn/showproblem.php?pid=2089" target="_blank" rel="noreferrer"&gt;题目链接&lt;/a&gt;
&lt;strong&gt;题意：问区间[n,m]中，不含数字4，也不含数字串“62”的所有数的个数。&lt;/strong&gt;&lt;/p&gt;</description></item><item><title>codeforces #341 div 2 E. Wet Shark and Blocks (数位dp+矩阵加速)</title><link>https://111qqz.com/en/post/acm-icpc/2016/2016-02-08-cf621e/</link><pubDate>Mon, 08 Feb 2016 09:04:41 +0000</pubDate><guid>https://111qqz.com/en/post/acm-icpc/2016/2016-02-08-cf621e/</guid><description>&lt;h3 class="relative group"&gt;&lt;a href="http://codeforces.com/problemset/problem/621/E" target="_blank" rel="noreferrer"&gt;http://codeforces.com/problemset/problem/621/E&lt;/a&gt;
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&lt;p&gt;题意：有b组数，每组数均有n个且相同。你必须在每组选一个数，组成一个新数sum，使得sum % x == k，问方案数 % (1e9+7)。&lt;/p&gt;</description></item></channel></rss>